We have vectors: $\vec v_{1} = \begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix}$ , $\vec v_{2} = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$, $\vec v_{3} = \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$ and they are eigenvectors of the matrix $A$ which corresponds to the eigenvalues: $\lambda_{1} = -2$, $\lambda_{2} = -1$, $\lambda_{3} = 2$, and let $\vec x =$ $\begin{bmatrix} -3 \\ 0 \\ -4 \end{bmatrix}$.
Express $\vec x$ as a linear combination of $\vec v_{1}, \vec v_{2}, \vec v_{3}$ and find $A\vec x$.
So I solved the linear combination part by performing $rref(A)$ and finding that the linear combination of the $\vec x$ is:
$\vec x = -2\vec v_{1} – 2\vec v_{2} -1 \vec v_{3}$.
Now I don't know how to solve the last bit of the problem, find $A \vec x$.
My thinking was $A$ is the eigenvalue and I could do the following:
$\lambda_{1} \vec v_{1} + \lambda_{2} \vec v_{2} + \lambda_{3} \vec v_{3} = A\vec x$. (I am incorrect.)
Any thoughts on how I could solve the last part? I thought A being an eigenvalue I could find $A\vec x$ the way I showed.
Best Answer
Note that
$Ax = A(-2v_1 - 2v_2 - 1v_3) = -2Av_1 -2Av_2 -1 Av_3 = -2 \lambda_1v_1 -2 \lambda_2 v_2 - \lambda_3 v_3 = 4v_1 + 2v_2 -2 v_3$
Then plug in $v_1, v_2, v_3$